Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the negative of the measured value if the polarity is not the same as the standard cell.) Take [Cu2+] to be 1 M. Show your calculations. Ecell=0.4249v, Ecell=-0.00191. M E =E- 0.0592/2 • log ([Cu2+]/[Ag+1?) (10) cell

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

Where E⁰= 0.4249V

E(Cell) = -(-0.0019V) -Measured value-

[Cu2+] = 1M

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

i would say b

explanation:

but i think i would need more information than that